HSC Arts 12th Board ExamMaharashtra State Board
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Properties of Inverse Trigonometric Functions

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Inverse of Sin, Inverse of cosin, Inverse of tan, Inverse of cot, Inverse of Sec, Inverse of Cosec

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We know that `sin"θ: R"→[-1/1]` and

 

`sin^-1θ: [-1,1]→ [-pi/2, pi/2]`


So, `sin(sin^-1)=x` then, `-1≤ x≤1` or `x∈ [-1,1]`


And, `sin^-1(sin x)=x` then, `x∈ [-pi/2, pi/2]`


1) `sin(sin^-1)=x, x∈ [-1,1]`


2) `sin^-1(sin)= x, x∈ [-pi/2, pi/2]`


3) Also `cosθ: "R"→[-1/1] , cos(cos^-1x)= x, x∈ [-1,1]`


if `x∉ [-1,1]`, then the answer cannot be determined.


4) `cos^-1(cos x)= x, x∈ [0, pi]`


But, if `x∉ [0, pi]`, we will transform x in such a way that x will belong to `[0, pi]`


5) `tan(tan^-1 x)= x, x∈ "R"`


6) `tan^-1(tan x)= x, x∈ (-pi/2, pi/2)`


But, if `x∉ (-pi/2, pi/2),` we will transform x in such way such that x will belong to `(-pi/2, pi/2)`


7) `cot (cot^-1 x)= x , x∈ "R"`


8) `cot^-1(cot x)= x, x∈ (0, pi)`


But if `x∉ (0, pi)`, we will transform x in such a way that it will belong to `(0, pi)`


9) `sec(sec^-1 x)= x, x∈ "R"- (-1,1)`


10) `sec^-1(sec x)= x, x∈ [0, pi]- {pi/2}`


But, if `x ∉[0, pi]- {pi/2},` we will transform x in such a way that it will belong to `[0,pi]- {pi/2}`


11) `cosec(cosec^-1x)= x, x∈"R"- (-1,1)`


12) `cosec^-1(cosec x)= x, x∈ [-pi/2, pi/2]- {0}`


But, if `x∉ [-pi/2, pi/2]- {0}`, we will transform x in such a way that it will belong to `[-pi/2, pi/2]- {0}`


While referring to this properties we must take this two points in consideration
i) Domain of each property will be intersection of principal branch of domain of all Inverse 
Trigonometric Function involved in property.
ii) Equal angle both side should belongs to same quadrant. 
Now we will prove some other properties of inverse trigonometric functions.


1) i) `sin^-1  1/x= cosec^-1 x, x ≥ 1 or x ≤ 1`


Let's see how is ti poved, `sin^-1  1/x= cosec^-1 x`


`cosec^-1 x= y`


`x= cosec   y`


`x= 1/sin y`


`sin y = 1/x`


`y= sin^-1  1/x`


`sin^-1  1/x= cosec^-1 x`


Likewise we can prove the other functions too.


ii) `cos^-1  1/x= sec^-1 x, x ≥ 1 or x ≤ -1`


iii) `tan ^-1  1/x= cot^-1 x, x > 0`


2) i) `sin^-1 (-x)= -sin^-1 x, x∈[-1,1]` or `|x| ≤ 1`


`sin^-1(-x)= y`


`sin y= -x`


`x= -sin y= sin(-y)`


`sin^-1 x= -sin^-1 (-x) ⇒ sin^-1(-x)= -sin^-1 x`


Likewise we can prove the other functions too.


ii) `cosec^-1(-x)= -cosec^-1 x, |x| ≥ 1` or `x ≤ 1` and `x ≥ 1`


iii) `tan^-1(-x)= -tan^-1 x, x ∈ "R"`


3) i)  `sec^-1(-x)= pi- sec^-1 x, |x| ≥ 1` 


`sec^-1 (-x)= y`


`sec y= -x ⇒ x= -sec y`


`x= sec (pi- y)`


`sec^-1 x= pi- sec^-1 (-x)`


`sec^-1 (-x)= pi- sec^-1`


Likewise we can prove the other functions too.


ii) `cos^-1 (-x)= pi- cos^-1 x, x ∈ [-1,1]`


iii) `cot^-1 (-x)= pi- cot^-1 x, x∈ "R"`


4) i) `sec^-1 x+ csoec^-1 x= pi/2, |x| ≥ 1`


`sec^-1 x+ cosec^-1 x= pi/2`


`sec^-1 x=y ⇒ x= sec y`


`x= cosec (pi/2 -y)`


`cosec^-1 x= pi/2 -sec^-1 x ⇒ sec^-1+cosec^-1 x= pi/2`


ii) `sin^-1 x+ cos^-1 x= pi/2, x∈ [-1,1]`


iii) `tan^-1 x+ cot^-1 x= pi/2, x∈ R`


5) i) `tan^-1 x+ tan^-1 y= tan^-1  (x+y)/(1-xy) , xy<1`


`tan(x+y)= (tan x+ tan y)/(1- tan x tan y)`


Let, `x= tan^-1 α and y= tan^-1 β`


`tan( tan^-1 α + tan^-1 β)= [tan (tan^-1 α) + tan (tan^-1 β)]/ [1- tan( tan^-1 α) tan(tan^-1 β)] `

                                                                
`(tan^-1 α+ tan^-1 β)= tan^-1  (α+β)/ (1- αβ)`


In the above result, if we replace `y` by `– y`, we get the second result and by replacing `y` by `x`, we get the third result as given below.


6) i) `2tan^-1 x= sin^-1  (2x)/(1+x^2), |x|≤ 1`


`sin 2x= (2 tan x)/ (1+ tan^2 x)`


Let `tan x= α`


`sin 2 tan^-1 α= (2α)/ (1+ α^2)`


`2 tan^-1 α= sin^-1  (2x)/(1+ α^2)`


ii) `2tan^-1 x= cos^-1  (1-x^2)/(1+x^2), x ≥ 0`


iii) `2tan^-1 x= tan^-1  (2x)/(1-x^2), -1`

 

Likewise we can prove the other functions too.

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